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Sets of perfect squares ending in 256

Source: 2012 AIME I Problem 10

March 16, 2012
modular arithmeticAMCAIME

Problem Statement

Let S\mathcal{S} be the set of all perfect squares whose rightmost three digits in base 1010 are 256256. Let T\mathcal{T} be the set of all numbers of the form x−2561000\frac{x-256}{1000}, where xx is in S\mathcal{S}. In other words, T\mathcal{T} is the set of numbers that result when the last three digits of each number in S\mathcal{S} are truncated. Find the remainder when the tenth smallest element of T\mathcal{T} is divided by 10001000.
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